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A person's leg has a mass of 14 kg, measures 1.0 m from the hip to the heel, and has a rotational inertia of 3.1 kg-m². If you treat the leg as a physical pendulum with small oscillations, what is its period of oscillation? Assume the center of mass of the leg is halfway down the leg.

A. 4.7 s

B. 0.21 s

C. 0.94 s

D. 1.5 s

E. 1.3 s

1 Answer

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Final answer:

To calculate the period of oscillation for the person's leg treated as a physical pendulum, the given formula is used, yielding a period of approximately 1.3 seconds, which corresponds to option E.

Step-by-step explanation:

To determine the period of oscillation for a person's leg treated as a physical pendulum, we can use the formula for the period of a physical pendulum:
T = 2π * √(I / (mgd))

Where:

  • T is the period of oscillation,
  • I is the moment of inertia,
  • m is the mass of the pendulum,
  • g is the acceleration due to gravity (approximately 9.8 m/s²), and
  • d is the distance from the pivot point to the center of mass.

Given that the person's leg has a mass (m) of 14 kg, a length from the hip to the heel of 1.0 m, and a rotational inertia (I) of 3.1 kg·m², with the center of mass halfway down the leg, we can plug these values into our formula as follows:
T = 2π * √(3.1 kg·m² / (14 kg * 9.8 m/s² * 0.5 m))

Solving for T using a calculator yields:

T ≈ 2π * √(3.1 / 68.6) ≈ 2π * √(0.04518) ≈ 2π * 0.2126 ≈ 1.337 s

Therefore, the period of oscillation is approximately 1.337 seconds, which rounds to 1.3 seconds. This corresponds to option E.

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